Chapter 11
Study Guide
-----------------Chi-Square
THE ANALYSIS OF
CATEGORICAL DATA AND
GOODNESS-OF-FlT TESTS
This section will explore inference for categorical data using chisquare test procedures. These procedures are used with univariate as
well as bivariate data sets if the variables are categorical.
OBJECTIVES
3 Carry out a chi-square goodness-of-fit test.
» Carry out chi-square tests for homogeneity of proportions and for
independence.
CHI-SQUARE TESTS FOR UNIVARIATE DATA
(Introduction to Statistics & Data Analysis 3rd ed. pages 647-€55/4th ed. pages 700-708)
In this section, we extend techniques for analyzing univariate
categorical data sets. Here we can consider questions about variables
that involve two or more categories. A chi-square test (y2} for
univariate data allows us to test hypotheses about the proportions
falling into the different categories for a categorical variable. We do
this by looking at the frequencies observed in each of the categories
and comparing them to what would be expected if a null hypothesis
were true. The hypothesized proportions can be equal or they can be
different for each category.
EXAMPLE A new donut shop plans to sell plain, strawberry, blueberry,
and cinnamon donuts. They wonder if there is a preference for one of
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Chapter 11
these types of donuts or if each type is preferred by the same
proportion of customers. If we let
pp = proportion
ps = proportion
pb = proportion
p_ = proportion
of customers preferring plain donuts
of customers preferring strawberry donuts
of customers preferring blueberry donuts
of customers preferring cinnamon donuts
We are interested in knowing if pp = ps = pb = pc = 0.25 or if there is
evidence that these proportions are not all the same. To answer this
question, a random sample of 60 customers is surveyed, and each
person in the sample is asked which of the four donut types they
prefer.
Instead of running a separate two-sample proportions test for each
of the possible pairs of proportions, a chi-square test will allow us to
decide if the proportions we have observed in our sample are
significantly different from the hypothesized proportions. The chisquare goodness-of-fit test will allow us to test the following
hypotheses:
Ha •. not all of the proportions are equal to 0.25
(For chi-square tests, it is acceptable to state hypotheses in words
on the AP Exam.)
To make the decision to reject or fail to reject the null hypothesis,
the chi-square test will compare the number of responses observed in
each category to what we would expect to see in each category if the
null hypothesis is true. If the difference is too large, we reject the null
hypothesis. Otherwise, we fail to reject, the null hypothesis.
Data on a single categorical variable is usually summarized using a
one-way frequency table. Returning to the previous example, suppose
that the sample of 60 customers resulted in the data summarized in the
table below. The table entries are observed frequencies or counts.
Types of Ponut Preferred
Observed Count
Plain
Strawberry
Blueberry
Cinnamon
13
12
16
19
These counts represent the number of times a person in the sample
selected each particular donut type. Expected counts are calculated
using the hypothesized proportions from the null hypothesis. In this
example, all four hypothesized proportions are equal to 0.25, so if the
null hypothesis is true, we would expect to see the same number
preferring each type. We calculate expected counts by multiplying the
sample size by each hypothesized proportion:
The Analysis of Categorical Data and Goodness-of-Fit Tests * 263
npp = 60(0.25) = 15
nps = 60(0.25) = 15
npb = 60(0.25) = 15
npc = 60(0.25) = 15
The expected counts can now be entered into the table as shown
here.
Types of Donuts Sold
Plain
Strawberry
Blueberry
Cinnamon
Observed
Count
13
12
16
19
Expected
Count
60(0.25) = 15
60(0.25) =15
60(0.25) = 15
60(0.25) = 15
Now we are ready to calculate the value of the test statistic. This test
statistic is called the goodness-of-fit statistic because it considers how
well the observed values fit to what we expected to see. Calculating the
value of a chi-square (%2} test statistic is relatively easy.
_ (observed cell count - expected cell count)2
expected cell count
2
which is often abbreviated
,(0 - E)2
For the example problem, the value of this test statistic is
2
(13-15)2 1 (12-15)2 (16-15)2
15
15
15
(-2) 2 ( C3)2 (I)2 (4)2
15 15
15
15
4 9 1 16
~15 + 15 + 15 15
30
15
=2
_
(19-15)2
15
To find the associated P-value, we use a chi-square distribution with
df = k - 1, where k is the number of categories of a categorical variable.
The P-value is the area to the right of the computed test statistic value
(the chi-square goodness-of-fit test is an upper tail test). This area is
found by referring to the chi-square table or by using a graphing
calculator or other technology.
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Chapter 11
Probabilityp
The chi-square table is similar to the t table. You locate the df in the
leftmost column then read across the row to find the value of the %2
test statistic. If the value is between two columns, then the P-value is
between the two corresponding tail probabilities.
Chi-Square Tail Probability (%2)
df
0.25
0.20
0.15
0.10
0.05
0.025
1
1.32
1.64
2.07
2.71
3.84
5.02
2
2.77
3.22
3.79
4.61
5.99
7.38
3
< 3Lll
4.64
5.32
6.25
7.81
9.35
5.99
6.74
7.78
9.49
11.14
Z
5.39 1
4
:
I^>
1
Since our xz = 2.00 and this value is smaller than 4.11, we know the
tail probability is greater than 0.25. Using any reasonable « level will
result in a failure to reject the null hypothesis.
The only things that we have not yet discussed are the overall
assumptions required for the chi-square test. The first assumption is
that we have a random sample of observations of a categorical
variable. In addition, the sample size must be large enough that the
following conditions are met:
1. No expected counts are < 1,
2. All of our expected counts should be > 5 and if they are not,
3. No more than 20% of the expected counts are < 5.
Returning to the donut example:
HYPOTHESIS
H0 : pp = 0.25, ps = 0.25, pb = 0.25, pc = 0.25
Ha : H0 is not true; in other words, not all of the proportions are equal
We will use a significance level of 0.05 for this test.
The Analysis of Categorical Data and Goodness-of-Fit Tests
TEST AND ASSUMPTIONS
Test: Chi-square goodness-of-fit test
Assumptions:
1. The sample was a random sample of customers
2. All expected counts are > 5, so the sample size is large enough.
TEST STATISTIC As seen earner, the calculated test statistic for this test
would be
2 _(13-15) 2 [
(12-15)2 | (16-15)2
15
15
15
_(-2) 2 [ (-3)2 (I)2 (4)2
15
15
15 15
(19-15)2
+
15
30
~15
=2
=
with df = 3
Based on df = 3 and %2 = 2, the P-value is greater than 0.25.
CONCLUSION Since P-value > a, there is insufficient evidence to reject
the null hypothesis. In other words, there is not convincing evidence
that the four types of donuts are not equally preferred.
AP Tip
For all chi-square tests, make sure to provide the expected counts
and verify that they are large enough.
SAMPLE PROBLEM 1 A music producer is interested in marketing a new
artist via ads in movie theaters. The target age group is teenagers from
14-18 years of age. The company developing the advertisement has
offered to run the ad in conjunction with the following types of
movies: 20% of the time with comedies, 50% with dramas, and 30%
with action films. However, the movie producer is not sure these
percentages reflect the types of movies that teens attend. To
investigate, each person in a random sample of teens was asked what
type of movie they had seen most recently, resulting in the following
data:
Movie Most Recently Watched By 1OO Teens
Comedy
Observed Count
41
Drama
35
Action
24
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Chapter 11
Do these data provide convincing evidence that the proportions of
teens watching the different types of movies are different than the
proportions proposed by the company developing the ad? Use a
significance level of 0.05.
SOLUTION TO PROBLEM 1
HYPOTHESIS
pc= proportion of teens who watched a comedy
pd= proportion of teens who watched a drama
pa= proportion of teens who watched an action films
H0:pc=0.2;
pd = 0.5; pa = 0.3
Ha : the null hypothesis is not true (at least one of the proportions of teens watching
the various film types is not equal to the hypothesized proportion).
a=0.05
ASSUMPTIONS The problem states the teens were a random sample.
All the expected counts are greater than or equal to 5 (see table
below).
movie most Recently watcnea B f luu Teens
Expected Counts
Comedy
Drama
Action
100(0.2) = 20
100(0.5) = 50
100(0.3) = 30
TEST STATISTIC
%z Goodness-of-fit test:
-20)2
(35-50)2 _ (24-3Q)2
20
(21)2
20
- -|
50
(-15)2
50
30
i
1- -
30
= 22.05 + 4.5 + 1.2
= 27.75
with df = 2
CONCLUSION The P-value associated with %2= 27.72 and df = 2 is
approximately 0.000. Since 0.000 < 0.05, there is strong evidence to
reject the null hypothesis. In other words, based on this sample there
is strong evidence that at least one of the proportions is not equal to
the proportion stated by the ad company.
Just as with other hypothesis tests, it is important to address all
parts of the test. Clearly conveying your understanding of the
procedure will include
» explaining the notation used,
a addressing all assumptions that are required for the test,
• demonstrating correct mechanics,
a and finally writing a conclusion in context.
The Analysis of Categorical Data and Goodness-of-Fit Tests *
As with other tests, the mechanics can be done using a graphing
calculator, but it is strongly recommended that you still show the
initial set-up. This lets the AP Reader know you understand how to
compute the value of the test statistic.
TESTS FOR HOMOGENEITY AND INDEPENDENCE
IN A TWO-WAY TABLE
(Introduction to Statistics & Data Analysis 3rd ed. pages 660-671/4th ed. pages 711-722)
A chi-square test procedure can also be used with bivariate categorical
data, which is usually displayed in a two-way frequency table. There
are two different types of investigations that arise from this type of
data. One type involves inferences about association between two
different categorical variables being observed on a single sample. The
other type involves comparing two or more populations or treatments
when a single categorical variable is observed. The calculation
procedure is the same for both types of investigation, but the primary
question of interest is different.
TEST FOR HOMOGENEITY
In a chi-square test for homogeneity, we are interested in whether the
proportions falling into each of the possible categories of a categorical
variable are the same for all of the treatments or populations studied.
In this case, the null hypothesis is that the distribution of the
categorical variable is the same for each population or treatment.
EXAMPLE A snack manufacturer produces three types of chips in two
different locations (Location A and Location B). Sometimes bags of
chips are damaged in the packaging process. Each bag in a random
sample of 45 bags of chips packaged at Location A and in a random
sample of 30 bags of chips packaged at Location B was classified into
one of three categories: no damage, minimal damage, and severe
damage. The resulting data is summarized in the table below. The
manufacturer was interested in determining if there was sufficient
evidence to conclude that the proportions falling into each of the three
damage categories is not the same for the two locations.
No
Damage
Minimal
Damage
Severe
Damage
Row Totals
Location A
15
18
12
45
Location B
8
12
10
30
23
30
22
75
Column Totals
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268 •:• Chapter 11
To answer the question of interest the hypotheses would be
H0: There is no difference in the proportions falling into each damage
category for the two locations
Ha: The proportions falling into each damage category are not the
same for the two locations
The assumptions that need to be are (1) the sample must be a
random sample and (2) all expected counts must be at least 5.
The expected counts for a chi-square test of homogeneity are
calculated using the following formula:
(row totalKcolumn total)
table total
For example, the expected count for the cell of no damage and
Location A is
(53)(23)
= 16.04
76
A way to record these expected counts is either in a separate twoway table of just the expected counts or in parenthesis beside each
observed count as shown in the table below. All of the expected counts
are greater than or equal to 5.
No Damage
Minimal
Damage
Severe
Damage
Row Totals
Location A
19 (16.04)
20 (18.13)
14 (18.83)
53
Location B
4 (6.96)
6 (7.87)
13 (8.17)
23
23
26
27
76
Column Totals
Degrees of freedom for the chi-square test of homogeneity is
computed as follows:
where r = number of rows and c = number of columns
Note that the total column and row are not counted in computing df.
The chi-square test statistic that was used in the goodness-of-fit test
is also used here.
%z test for homogeneity
2 _(19-16.04) 2
(20-18.13)2 (14-18.83)2 (4-6.96)2 { (6-7.S7)2 : (13-8.17)2
X ~
16.04 +
18.13
+
18.83 + 6.96 + 7.87
8.17
= 6.5
withdf = (2-l)(3-l) = 2
The Analysis of Categorical Data and Goodness-of-Fit Tests •* 269
Using the graphing calculator technology, you first enter a matrix
of the observed counts and then run a chi-square test from the test
menu. The calculator will calculate the expected counts and store them
in matrix[B].
MflTRIXIR] 2 x3
[ 19
[ H
EO
fi
1H
13
]
]
*2-Test
Observed : [fi]
Expected: [B]
Calculate Draw
X 2 -Test
X*=6. 534131993
p=. 0381181018
df=2
Finally, since P < a (0.038 < 0.05), we reject the null hypothesis.
There is strong evidence that the proportions falling into each of the
three damage categories is not the same for both locations.
TEST FOR INDEPENDENCE
The chi-square test for independence, also known as the chi-square
test for association, is used to investigate if there is an association
between two categorical variables. The calculations will proceed in the
same manner as the test for homogeneity; however, we are actually
looking to see if knowing the value of one variable provides
information about the value of the other variable.
EXAMPLE A car manufacturer has two production lines building three
types of cars. An engineer is wondering if there is an association
between the type of car and the production line that made the vehicle
for cars that are found to have major defects. Each car in a random
sample of 75 cars selected from all cars found to have major defects
was classified according to the type of car and the production line that
produced the car. The resulting data is given in the table below.
Sedan
Wagon
Truck
Line A
13
9
12
Line B
18
12
11
The question of interest is whether there is an association between
car type and production line for cars with major defects.
To answer this question, we use the chi-square test for independence.
HYPOTHESIS
H0 : Production line and car type are independent.
Ha : Production line and car type are not independent.
a = 0.05
ASSUMPTIONS The sample was a random sample of cars with major
defects.
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Chapter 11
The expected counts are all greater than or equal to 5, so we can
proceed (see table below).
Expected Counts for Auto Errors
Sedan
Wagon
Truck
Line A
14.1
9.5
10.4
LineB
16.9
11.5
12.6
TEST STATISTIC Since this is a %2 test for independence, the calculations
would be
2
^(13-14.1)2 | (9-9.S)2 | (12-10.4)2
14.1
9.5
10.4
[
(18-16.9)2 | (12-11.5)2
16.9
11.5
<
(11-12.6)2
12.6
df = 2
P = 0.73
CONCLUSION Since 0.73 is not smaller than 0.05, we fail to reject the nuU
hypothesis. Therefore, we do not have sufficient evidence of an
association between car type and production line.
Notice, the steps in the chi-square hypothesis tests are the same as
for all other tests: State hypotheses, identify the test by name or
formula and check assumptions required for the test, calculate the
value of the test statistic and the associated P-value, and write a
conclusion in context that is linked to the P-value.
INTERPRETING AND COMMUNICATING THE RESULTS OF
STATISTICAL ANALYSES
(Introduction to Statistics & Data Analysis 3rd ed. pages 677-680/4th ed. pages 727-730)
This particular set of tests involved writing our conclusions in terms
that match the setting of the individual test. Each chi-square test is
performed in a different setting depending on the data we have to
work with. It is important to state which of the three chi-square tests
you are using when carrying out one of these tests. The goodness-offit, homogeneity, and independence tests all have different hypotheses.
Be clear which test you are using.
CATEGORICAL DATA AND GQODNESS-OF-FIT TESTS:
STUDENT OBJECTIVES FOR THE AP EXAM
•
You will be able to determine degrees of freedom for chi-square
tests of goodness-of-fit, homogeneity, and independence.
^ You will be able to calculate expected counts for chi-square tests of
goodness-of-fit, homogeneity, and independence.
• You will be able to carry out chi-square tests of goodness-of-fit,
homogeneity, and independence.
The Analysis of Categorical Data and Goodness-of-Fit Tests *
«
You will identify the correct hypothesis for each of the three chisquare scenarios.
« You will be able to interpret conclusions in contest for chi-square
tests of goodness-of-fit, homogeneity, and independence.
MULTIPLE-CHOICE QUESTIONS
1. Each car in a random sample of 200 cars sold in 2010 at a large car
dealership was classified by color. Sixty of the cars were white, 80
were blue, 20 were silver and 10 were red. What is the value of the
chi-square statistic in a test of H0 : pw = pb = ps = pr = 0.25 ?
(A) 18
(B) 32
(C) 50
(D)70
(E) 110
2. Which of the following is not true of the %2 probability
distribution?
(A) For small degrees of freedom, the distribution is right skewed.
(B) As the degrees of freedom increase, the %2 distribution
approaches a normal distribution.
(C) All of the area under a chi-square curve is associated with
positive values.
(D) The total area under the^2 curve is equal to one.
(E) The mean of a chi-square distribution is 0.
3. There are four surgical methods currently being used to place
medical implants in patients. After surgery, patients are monitored
and their pain level is recorded as severe, moderate, or mild. What
are the degrees of freedom for a test to determine if there is an
association between surgical method and pain level?
(A) 2
(B) 3
(C)4
(D)6
(E) 12
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Chapter 11
Questions 4-7 refer to the following set of data:
A group of AP Statistics students wanted to see if plain, peanut and
almond M&Ms have the same color distribution. To test this, students
took a random sample of each type of M&M and classified the candies
in the sample by color. They plan to carry out a chi-square test to
decide if there is evidence that the color distributions are not the same
for the three types of M&Ms.
Red
Blue
Yellow
20
18
15
Peanut
8
6
Almond
7
11
Plain
Orange
Brown
10
14
12
8
25
5
7
10
12
10
9
Green
4. What is the expected count for green peanut M&Ms?
25
(A)
47
25x47
(B)
47
59x47
(C)
59
59x47
(D)
207
25x47
IE)
207
5. What are the correct degrees of freedom for the appropriate test?
(A) 206
(B) 18
(C) 10
CD) 7
(E) 4
6. What is an appropriate set of hypotheses?
(A) H0: There is no association between the type of M&M and
color.
Ha: There is an association between the type of M&M and
color.
(B) H0: There is an association between the type of M&M and
color.
Ha: There is no association between the type of M&M and
color.
(C) H0: The color proportions are the same for all 3 types of M&Ms.
Ha: The color proportions are not the same for all 3 types of
M&Ms.
(D) H0: Tne proportions of red M&Ms are the same for all 3 types
of M&Ms.
Ha: The proportions of red M&Ms are not the same for all 3
types of M&Ms.
(E) Ho: There is no difference in the type of M&M for green M&Ms.
Ha: There is a difference in the type of M&M for green M&Ms.
The Analysis of Categorical Data and Goodness-of-Fit Tests *• 273
7. The value of the chi-square test statistic is 22.9. What conclusion
would be reached in a test of the hypotheses of interest using a
significance level of 0.05?
(A) Fail to reject H0 and conclude there is strong evidence of an
association.
(B) Fail to reject H0 and conclude there is strong evidence that
the color distributions are the same.
(C) Fail to reject H0 and conclude there is not strong evidence
that the color distributions are the same.
(D) Reject H0 and conclude there is strong evidence of an
association.
(E) Reject H0 and conclude there is strong evidence that the color
distributions are not the same.
8. A cupcake store manager believes that half of the cupcakes sold
are chocolate and that the other half is equally divided between
vanilla and cherry. If the manager's belief is correct and 300
cupcakes are sold, how many would you expect to be cherry
cupcakes?
(A) 300
(B) 200
(C) 150
(D) 100
(E) 75
9. Each person in a random sample of patrons of a local mall was
surveyed regarding a public smoking area outside one of the mall
entrances. Each person was asked if they approved of the idea of a
public smoking area in the mall. The resulting data is summarized
in the table below. The mall management would like to know if
there is a relationship between gender and approval of the
smoking area. What would be an appropriate set of hypotheses?
Public Smokers
Approve
Do Not Approve
Males
28
57
Females
39
31
(A) H0: Gender and approval are not independent.
Ha: Gender and approval are independent.
(B) H0: Gender and approval are independent.
Ha: Gender and approval are not independent.
(C) H0: Knowing a person does not approve of a public smoking
area indicates their gender.
Ha: Knowing a person does not approve of a public smoking
area does not indicate their gender.
(D) H0: There is no difference in gender distribution based on
approval.
Ha: There is a difference in gender distribution based on
approval.
(E) H0: There is an association between gender and approval.
Ha: There is no association between gender and approval.
'
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Chapter 11
10. A local bagel shop makes six types of bagels and eight types of
cream cheese toppings. Suppose that all customers order one type
of bagel and one type of cream cheese. If customers are classified
according to the type of bagel and type of cream cheese, how
many degrees of freedom would be appropriate for a test to
determine if there is an association between type of bagel and type
of cream cheese purchased?
(A) 7
(B) 12
(C) 14
(D) 35
(E) 48
11. In a recent opinion poll, likely voters were asked if they would
continue to support health care reform if the cost per taxpaying
citizen was increased by $800 per year. Political preference was
also recorded. The P-value for a chi-square test of independence is
0.001. Which of the following is a correct interpretation of this
result?
Democrat
Republican
Independent
Libertarian
Support
125
87
99
48
Don't Support
75
113
101
52
(A) Since P = 0.001, there is strong evidence that political
preference and support are independent of each other. In other
words, knowing their political preference gives no insight into
their support of the increase.
(B) Since P = 0.001, there is strong evidence that political
preference and support are not independent of each other. In
other words, knowing their political preference gives insight
into their support of the increase.
(C) Since P = 0.001, there is strong evidence that political
preference and support are independent of each other. Since
they are independent, there is no association between the two
variables.
(D) Since P = 0.001, there is insufficient evidence that political
preference and support are independent of each other. Since
they are not independent, there is no association between the
two variables.
(E) Since P = 0.001, there is insufficient evidence that political
preference and support are not independent of each other. In
this case, knowing their political preference would not help in
knowing their support.
The Analysis of Categorical Data and Goodness-of-Fit Tests
12. Two different cereal companies each make a bran cereal, a corn
cereal, and a rice cereal. Each person in a random sample of 300
potential customers tasted all six cereals and selected their
favorite. The choice was then classified according to type of cereal
and the company that made the cereal. The data is shown in the
table below. What type of test would be appropriate?
Cereal 1
Cereal 2
Cereal 3
Company A
51
48
47
Company B
43
65
46
(A) Chi-sguare goodness-of-fit test
(B) Chi-sguare test of independence
(C) Chi-sguare test of homogeneity
(D) Either a chi-sguare test for independence or homogeneity is
appropriate
(E) A chi-square test is not appropriate in this situation
13. Which is not true for a chi-sguare test of independence?
I. The test is based on bivariate categorical data.
II. df is calculated using k - 1, where k = the number of categories.
III. The test is an upper-tailed test.
(A) I, II, and III
(B) I and II
(C) II and HI
(D) I only
(E) II only
Questions 14-15 refer to the following set of data:
A movie theater recorded the type of snack item purchased and the
type of movie the customer was attending for each person in a random
sample of theater customers who made a snack bar purchase. The
resulting data is given in the table below.
Soda
Popcorn
Candy
Children's Movie
70
83
47
Action Movie
61
49
20
14. If the manager wanted to determine if there is an association
between the type of movie and type of snack sold, which type test
would be appropriate?
(A) Chi-sguare goodness-of-fit
(B) Chi-sguare test for homogeneity
(C) Chi-sguare test for independence
(D) Two-sample proportions test
(E) Multiple sample proportions test
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Chapter 11
15. Using the data in the table above, the value of the chi-square
statistic is 5.66. Which of the following is a correct statement about
the P-value for this test?
(A) P-value > 0.10
(B) 0.05 < P-value < 0.10
(C) 0.01 < P-value < 0.05
(D) 0.001 < P-value < 0.01
(E) P-value < 0.001
FREE-RESPONSE PROBLEMS
1. Regina is worried that the color of her new cardigan will attract
the attention of killer bees in southern California where she is
going to hike. To settle her nerves she looks at the American Killer
Bee Association website. It shows that these bees are highly
agitated by various colors. They have found that 75% of bees are
agitated by green, 9% by blue, 6% by purple or pink, and the
remaining 10% by other colors.
(a) In a random sample of 200 killer bees, how many would you
expect to be agitated by each color?
(b) A recent study of 120 randomly selected people stung by killer
bees last year found the individuals were wearing the colors
shown in the table below. Do these data provide convincing
evidence that the color distribution of colors worn by people
stung by killer bees is different from the percentages given on
the website?
Color Worn By Individual Stung By Killer Bees
Green
Purple/Pink
Other
Blue
86
21
6
7
2. A restaurant offers both dine-in and take-out service. Customers
can pay for their meal in cash, by credit card or by debit card. The
restaurant owner wonders if there is an association between the
method of payment and the type of service. To investigate, a
random sample is selected from the orders placed during the last
year and the method of payment and the type of service is
recorded for each of these orders. The data is summarized in the
table below.
Cash
Credit
Debit
Dine-in
34
122
32
Take-out
70
95
47
(a) Should the restaurant owner carry out a test of homogeneity or
a test of independence to answer his question?
(b) Carry out a test to answer the question of interest to the
owner. Use a significance level of 0.05 for your test.
The Analysis of Categorical Data and Goodness-of-Fit Tests »> 277
Answers
MULTIPLE-CHOICE QUESTIONS
1. D. The calculation would be found as follows:
(Introduction to Statistics & Data Analysis 3rd ed. pages 647656/4th ed. pages 700-708).
2. E. The x2 distribution is right skewed with all of its area
associated with positive values. The mean of a chi-square
distribution can't be 0 (Introduction to Statistics & Data Analysis
3rd ed. pages 647-656/4th ed. pages 700-708).
3. D. There are four surgical methods and three pain levels.
d f = ( r - l ) ( c - l ) = ( 4 - l ) ( 3 - l ) = 6 (Introduction to Statistics &
Data Analysis 3rd ed. pages 660-671/4th ed. pages 711-722).
4. D. The expected counts in a two-way table are found using
(row total x column total) ,r .
„.
- (Introduction to Statistics & Data
table total
Analysis 3rd ed. pages 660-671/4th ed. pages 711-722).
5. C. In a two-way table, degrees of freedom are calculated by
(r - l)(c - 1) = (3 - 1)(6 - 1) = 10 (Introduction to Statistics & Data
Analysis 3rd ed. pages 667-671/4th ed. pages 711-722).
6. C. This is a test of homogeneity. Three populations are being
compared on the basis of a categorical variable (color)
(Introduction to Statistics & Data Analysis 3rd ed. pages 660671/4th ed. pages 711-722).
7. E. x2 = 22.9 and df = 10, gives a P-value = 0.01 which is less than
0.05. This means there is strong evidence to conclude that the color
distributions are not all the same (Introduction to Statistics & Data
Analysis 3rd ed. pages 660-671/4th ed. pages 711-722).
8. E. The expected count is calculated is 300(0.25) = 75 (Introduction
to Statistics & Data Analysis 3rd ed. pages 647-656/4th ed. pages
700-708).
9. B. A test of independence between the variables would be used to
answer the question of interest (Introduction to Statistics & Data
Analysis 3rd ed. pages 660-671/4th ed. pages 711-722).
10. D. (6 - 1)(8 - 1) = 5(7) = 35 (Introduction to Statistics & Data Analysis
3rd ed. pages 660-671/4th ed. pages 711-722).
278 *
Chapter 11
11. B. The P-value is smaller than the significance level, so the null
hypothesis of independence would be rejected. This means there is
an association and knowing people's political party provides
information about their support of this issue (Introduction to
Statistics & Data Analysis 3rd ed. pages 660-671/4th ed. pages 711722).
12. B. Because a single sample is classified on the basis of two
categorical variables, the appropriate test is a test for
independence (Introduction to Statistics & Data Analysis 3rd ed.
pages 660-671/4th ed. pages 711-722).
13. E. The degrees of freedom for a goodness-of-fit test is calculated
using k-1. However, for a chi-square test of independence,
degrees of freedom is calculated using (r - l)(c - 1) (Introduction to
Statistics & Data Analysis 3rd ed. pages 660-671/4th ed. pages 711722).
14. C. This is a test of independence since we are trying to see if
knowing the type of movie provides any information about type of
snack purchased (Introduction to Statistics & Data Analysis 3rd ed.
pages 660-671/4th ed. pages 711-722).
15. B. With df = 2, the P-value is 0.059 (Introduction to Statistics & Data
Analysis 3rd ed. pages 660-671/4th ed. pages 711-722).
FREE-RESPONSE PROBLEMS
1. (a) The expected counts by color for 200 bees would be:
Green: 200(0.75) = 150,
Blue: 200(0.09) = 18,
Purple/Pink: 200(0.06) = 12,
Other: 200(0.10) = 20.
(b) Hypotheses
H0 : The proportions of bee stings by color is as specified on
the web site (pgreen = 0.75, pwue = 0.09, ppuvle/pmk = 0.06, pother = 0.10)
Ha : At least one of the color proportions is different from
what is specified inthe null hypothesis
a = 0.05
Test: Chi-square goodness-of-fit test
Assumptions
1. The problem states that the individuals in the sample were
randomly selected.
2. The expected counts are shown in the table below (note
that the sample size is 150, not 200 as in part (a). All
expected counts are greater than 5, so the sample size is
large enough.
The Analysis of Categorical Data and Goodness-of-Fit Te5t5 * 279
Expected Counts for 15O Stings
Green
Blue
Purple/Pink
Other
(90.0)
(10.8)
(7.2)
(12.0)
Test Statistic
X2 = 12.09, df = 3, P = 0.007
Conclusion
Since 0.007 < 0.05, there is strong evidence that to reject the
null hypothesis. Therefore, there is reason to say the
proportions of bee stings by killer bees are not all the same as
reported on the website
(Introduction to Statistics & Data Analysis 3rd ed. pages 647656/4th ed. pages 700-708).
2. (a) This is a test of independence. There was only one sample and
each individual in the sample was classified according to two
categorical variables,
(b)
Hypotheses
H0 : method of payment and order type are independent.
Ha : method of payment and order type are not independent.
a = 0.05
Assumptions
(1)
The sample was a random sample of orders.
(2)
The expected counts are given in the table below. All
expected counts are greater than 5, so the sample size is large
enough.
Expected Counts
Cash
Credit
Debit
Pine-in
$48.88
$101.99
$37.13
Take-out
$55.12
$115.01
$41.87
Test Statistic
(34-48.88) (122-101.99)2
+ ... = 17.291, withdf = 2
101.99
48.88
Conclusion
With a test statistic so extreme, our P-value = 0.000. This is
smaller than any reasonable significance level, so we reject the
null hypothesis. There is strong evidence that there is an
association between method of payment and order type.
2
X =
(Introduction to Statistics & Data Analysis 3rd ed. pages 660671/4th ed. pages 711-722).